proof of Banach-Steinhaus theorem
ProofOfBanachSteinhausTheorem
2004-11-12 07:55:39
2006-06-28 04:39:31
Proof
Banach-Steinhaus theorem
0
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Let
$$E_n = \{x\in X: \|T(x)\|\leq n\textnormal{ for all }T\in \mathcal{F}\}.$$
From the hypothesis, we have that
$$\bigcup_{n=1}^\infty E_n = X.$$
Also, each $E_n$ is closed, since it can be written as
$$E_n = \bigcap_{T\in\mathcal{F}}{T^{-1}(B(0,n))},$$
where $B(0,n)$ is the closed ball centered at $0$ with radius $n$ in $Y$,
and each of the sets in the intersection is closed due to the continuity of the operators.
Now since $X$ is a Banach space, Baire's category theorem
implies that there exists $n$ such that $E_n$ has
nonempty interior. So there is $x_0\in E_n$ and $r>0$ such
that $B(x_0,r)\subset E_n$. Thus if $\|x\|\leq r$, we have
$$\|T(x)\|-\|T(x_0)\|\leq \|T(x_0)+T(x)\|=\|T(x_0+x)\|\leq n$$
for each $T\in \mathcal{F}$, and so
$$\|T(x)\|\leq n+\|T(x_0)\|$$
so if $\|x\|\leq 1$, we have
$$\|T(x)\|= \frac{1}{r}\|T(rx)\| \leq \frac{1}{r}\left(n+\|T(x_0)\|\right) = c,$$
and this means that
$$\|T\| = \sup\{\|Tx\|: \|x\|\leq 1\} \leq c$$
for all $T\in\mathcal{F}$.